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2020 AMC 10A Problem 9

2020 AMC 10A Problem 9AMC 10 Step by Step

Name: Date: October 28, 2026

Divisibility & Factors · difficulty 2 of 5 · about 3 min · position P1-10

A single bench section at a school event can hold either $7$ adults or $11$ children. When $N$ bench sections are connected end to end, an equal number of adults and children seated together will occupy all the bench space. What is the least possible positive integer value of $N?$

A) $9$
B) $18$
C) $27$
D) $36$
E) $77$

Answer

Key insight

Cut a bench into 77 pieces: an adult uses 11 and a child 7, so each adult-child pair uses 18. So 77 times N must be a multiple of 18.

Solution

Adults and children take up different amounts of room, so first find a unit that measures both in whole numbers.

Cut each bench section into $77$ equal pieces (because $77 = 7 \times 11$ ).
$7$ adults fill a section, so one adult uses $77 \div 7 = 11$ pieces. $11$ children fill a section, so one child uses $77 \div 11 = 7$ pieces.
The adults and children come in equal numbers, so think of them as adult-child pairs. One pair uses $11 + 7 = 18$ pieces.
$N$ sections have $77N$ pieces. They are filled exactly by pairs, so $77N$ must be a multiple of $18$ .
Now $18 = 2 \times 3 \times 3$ and $77 = 7 \times 11$ share no common factor. So the $77$ does not help at all, and all of the $18$ must come from $N$ . The smallest choice is $N = 18$ .
Check: $18$ sections have $77 \times 18$ pieces, which is exactly $77$ pairs. That is $77$ adults and $77$ children. Adults fill $77 \div 7 = 11$ sections and children fill $77 \div 11 = 7$ sections, and $11 + 7 = 18$ .

The answer is $\boxed{\textbf{(B)}\ 18}$ .

Why this works

Choosing a small unit that both "sizes" divide evenly turns fractions of a bench into whole numbers. The question then becomes a divisibility question. If a number divides $77N$ but shares no factor with $77$ , it must divide $N$ itself.

Alternative approach

Test the choices from smallest. $N = 9$ : $9 \times 77 = 693$ pieces, and $693 \div 18 = 38.5$ , not a whole number of pairs. $N = 18$ : $18 \times 77 \div 18 = 77$ pairs, which works. So $18$ is the least.

The trap

Answering 77 because 7 times 11 feels natural, or answering 9 without checking that the adults and children come out to whole numbers.

Concise solution

Let $x$ be the number of adults, which equals the number of children. Each adult takes $\frac{1}{7}$ of a section and each child $\frac{1}{11}$ , so filling $N$ sections exactly means
$\frac{x}{7} + \frac{x}{11} = N \quad\Longrightarrow\quad 18x = 77N.$
The left side is a multiple of $18$ , so $77N$ must be as well. Because $77 = 7 \cdot 11$ shares no factor with $18 = 2 \cdot 3^2$ , the factor $18$ must divide $N$ itself. The least positive such $N$ is $18$ , achieved with $x = 77$ adults and $77$ children.

The answer is $\boxed{\textbf{(B)}\ 18}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 1: Integers: The Basics (§1.4, §1.5)
Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Number Theory, Chapter 3: Multiples and Divisors (§3.3, §3.4, §3.7)
Introduction to Number Theory, Chapter 5: Divisor Problems (§5.2, §5.3)

Techniques

Set up the equation/formula and compute; no special trick needed